I know that similar questions have been asked on this site before, but I haven't been able to find the answer to my specific question.

I want to show that the Noether charge defined in Lagrangian formalism generates corresponding symmetries. More, precisely:

Assume we have a Lagrangian $L(q(t),\dot q(t))$. Let us assume that, under an infinitesimal transformation $$\delta q(t) = \eta(q(t),\dot q(t),t), \ \delta \dot q (t) = \frac{d}{dt}\eta(q(t),\dot q(t),t)\tag{1}$$ the change of Lagrangian is given as: $$\delta L = \frac{d}{dt}K(q(t),\dot q(t),t)\tag{2}$$ Then the Noether charge $Q(q(t),\dot q(t),t)$ is defined by: $$Q(q(t),\dot q(t),t) := \frac{\partial}{\partial \dot q(t)}L(q(t),\dot q(t))\ \eta(q(t),\dot q(t),t) - K(q(t),\dot q(t),t)\tag{3}$$ The claim is that, if the define canonical momentum by $p(q,\dot q)= \frac{\partial}{\partial \dot q}L(q,\dot q)$: $$\delta q(t) = \left(\frac{\partial}{\partial p(t)}Q(q(t),\dot q(t),t)\right)_{q(t)},\qquad \delta p(t) = -\left(\frac{\partial}{\partial q(t)}Q(q(t),\dot q(t),t)\right)_{p(t)}\tag{4}$$ I was able to derive the first of above relations as follows. From, $$\delta L = \frac{\partial}{\partial q}L \delta q(t) + p(t) \delta \dot q(t)\tag{5},$$ by integrating by parts, we obtain: $$ \frac{d}{dt}Q = \delta q \left(\dot p - \frac{\partial}{\partial q}L \right) \\ \quad \quad \quad \quad \quad \ \ =\delta q \left(\frac{\partial p}{\partial q}\dot q + \frac{\partial p}{\partial \dot q}\ddot q - \frac{\partial}{\partial q}L \right) \\ \quad \quad \quad \ = \frac{\partial Q}{\partial t} + \frac{\partial Q}{\partial q}\dot q + \frac{\partial Q}{\partial \dot q}\ddot q\tag{6}$$ By equating terms proportional to $\ddot q$, we obtain: $$ \frac{\partial Q}{\partial \dot q} = \left( \frac{\partial p}{\partial \dot q} \right)_{q} \delta q \tag{7}$$ Then: $$ \left(\frac{\partial Q}{\partial p}\right)_{q} = \frac{\partial Q}{\partial \dot q} \left(\frac{\partial \dot q}{\partial p}\right)_q = \left( \frac{\partial p}{\partial \dot q} \right)_{q} \left(\frac{\partial \dot q}{\partial p}\right)_q \delta q = \delta q\tag{8}$$ My question is how we can show that $$\delta p(t) = -\left(\frac{\partial}{\partial q(t)}Q(q(t),\dot q(t),t)\right)_{p(t)}\tag{9}$$


It seems that it is required to assume equations of motion to obtain above identity. Consider the identity (obtained after removing the terms that include $\ddot q$ in $\frac{dQ}{dt} = \cdots$): $$ \delta q \left(\frac{\partial p}{\partial q}\dot q - \frac{\partial}{\partial q}L \right) = \frac{\partial Q}{\partial t} + \frac{\partial Q}{\partial q}\dot q \tag{10}$$ Now take partial derivative with respect to $\dot q$, and use commutativity of partial derivatives and the identity found above for $\frac{\partial Q}{\partial \dot q}$ to obtain: $$\frac{\partial Q}{\partial q} = \left(\dot q \partial_q p - \partial_q L \right)\partial_{\dot q}\delta q - \partial_t \delta q\partial_{\dot q} p - \dot q\partial_{\dot q}p \partial_q \delta q \tag{11}$$ Now, using $$\left(\frac{\partial}{\partial q}\right)_p = \frac{\partial}{\partial q} + \left(\frac{\partial \dot q}{\partial q}\right)_p\tag{12}$$ and $$\left(\frac{\partial \dot q}{\partial q}\right)_p = -\frac{\partial_{q}p}{\partial_{\dot q} p}\tag{13}$$ we find: $$\left(\frac{\partial Q}{\partial q}\right)_p = - \partial_q p \delta q - \partial_{\dot q}p \left(\dot q\partial_q \delta q + \partial_t \delta q \right) + \left(\dot q \partial_q p - \partial_q L \right) \tag{14}$$ If we assume equations of motion, that is: $$\partial_q L = \frac{d}{dt}p = \dot q \partial_q p + \ddot q \partial_{\dot q} p\tag{15}$$ We obtain: $$\left(\frac{\partial Q}{\partial q}\right)_p = - \partial_q p \delta q - \partial_{\dot q}p \left(\ddot q \partial_{\dot q} \delta q + \dot q\partial_q \delta q + \partial_t \delta q \right) = -\delta p \tag{16}$$ However, I can't see the physical meaning of this assumption.


1 Answer 1

  1. In this answer let us for simplicity restrict to the case of a regular Legendre transformation in a point mechanical setting, cf. this related Phys.SE post. (Generalizations to field theory and gauge theory are in principle possible, with appropriate modifications of conclusions.)

    Within the Lagrangian formalism, let there be given an infinitesimal vertical quasi-symmetry of the form $$\delta q^i ~=~y^i(q,\dot{q},t)\varepsilon, \qquad i~\in~\{1,\ldots,n\}, \qquad\delta t~=~0. \tag{A} $$ Noether's theorem then yields a corresponding Noether charge $Q_L(q,\dot{q},t)$ that is conserved on-shell.

  2. OP seems to essentially be asking the following question.

    Does the Noether charge $Q_L(q,\dot{q},t)$ generate the quasi-symmetry (A)?

    The answer is: Yes, the quasi-symmetry $$y^i(q,\dot{q},t)~=~\{q^i(t),Q_L(t)\}, \qquad i~\in~\{1,\ldots,n\},\tag{B} $$ is generated via the Peierls bracket $$\{ F,G \}~:=~\iint\!dt~dt^{\prime}~\sum_{i,k=1}^{n} \frac{\delta F }{\delta q^i(t)}~G^{ik}_{\rm ret}(t,t^{\prime})~\frac{\delta G }{\delta q^k(t^{\prime})} - (F\leftrightarrow G),\tag{C} $$ where $G^{ik}_{\rm ret}(t,t^{\prime})$ is the retarded Green's function, see e.g. various textbooks by Bryce S. DeWitt and this Phys.SE answer by user AccidentalFourierTransform. Unfortunately, within the Lagrangian formalism, we do not know the explicit form of the retarded Green's function $G^{ik}_{\rm ret}(t,t^{\prime})$, except in special cases. This makes OP's purely Lagrangian endeavour challenging & calculations cumbersome.

  3. However, we know that there is a bijective correspondence to a corresponding Hamiltonian formulation via a Legendre transformation, cf. e.g. this Phys.SE post. Moreover, there exists a bijective correspondence between conserved quantities $$Q_L(q,\dot{q},t) ~\approx~Q_H(q,p,t)\tag{D}$$ in the Lagrangian and the Hamiltonian formulations, cf. this Phys.SE post. A similar bijective correspondence exists for the corresponding quasi-symmetries. Therefore everything can be translated to the corresponding Hamiltonian formulation. We shall apply this strategy in this answer.

  4. In the Hamiltonian setting, OP's question becomes Statement 1 in my Phys.SE answer here, i.e. $$Y^I(z,t)~=~\{z^I(t),Q_H(t)\},\qquad I~\in~\{1,\ldots,2n\} .\tag{E} $$ Within the Hamiltonian formalism, the retarded Green's function $$G^{IK}_{\rm ret}(t,t^{\prime}) ~\simeq~\omega^{IK}~\theta(t\!-\!t^{\prime}),\qquad I,K~\in~\{1,\ldots,2n\},\tag{F} $$ is known explicitly in the adiabatic limit (where the Hamiltonian $H$ can be ignored). Here $\omega^{IK}$ denotes the symplectic unit. Therefore, the equal-time Peierls bracket becomes the canonical Poisson bracket.

  • $\begingroup$ I will learn more about Peierls bracket. However, I still can't understand what is wrong with following logic. Calculate $Q_L(q,\dot q,t)$ and calculate $p(t)$ as a function of $q(t)$ and $\dot q(t)$. Then calculate partial derivate of $Q_L$ with respect to $p$ keeping $q$ constant. This should give $\delta p(t)$. What I have shown in my post is that, this is only true if we assume equations of motion. $\endgroup$
    – Bronsteinx
    Sep 6, 2018 at 15:55
  • $\begingroup$ Comments: 1. Consider to explain in more detail your formula (4). 2. $\delta p$ seems irrelevant from a Lagrangian point of view. $\endgroup$
    – Qmechanic
    Sep 6, 2018 at 16:16
  • $\begingroup$ If we know $p$ as a function of $q$ and $\dot q$, then we can calculate variation of $p$ simply by $\delta p = \delta q \partial_q p + \delta \dot q \partial_{\dot q} p$. Now, if $Q_L = Q_H$, we know that $\delta p = \{p,Q_H\} = \{p,Q_L\} = - \left(\frac{\partial Q_L}{\partial q}\right)_p $. I wasn't able to show this without assuming equations of motion. I have no problem with showing $\delta q = \{q,Q_L\} = \left(\frac{\partial Q_L}{\partial p}\right)_q$ which is what I have done in $(8)$. $\endgroup$
    – Bronsteinx
    Sep 6, 2018 at 22:08
  • $\begingroup$ The reason why I am asking this question is that, passing from Lagrangian formalism to Hamiltonian formalism is not hard and all of this discussion in Hamiltonian formalism is also not hard therefore one should be able show that $Q_L = Q_H$ generates symmetries without too much work. It is possible that I am missing something about the correspondence between Lagrangian formalism and Hamiltonian formalism, that is what I am trying to understand. $\endgroup$
    – Bronsteinx
    Sep 6, 2018 at 22:08
  • $\begingroup$ Question: From symmetry in time, we can derive energy as a Noether charge, which is same as Hamiltonian. Now, for free particle, $\delta p = \epsilon m a \neq \{p,H\}$ offshell. I think this is the simplest example for my question. $\endgroup$
    – Bronsteinx
    Sep 8, 2018 at 6:41

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